A beam of charged particle enters in a region of magnetic field of 5 × 10^(−3) weber m and electric field of 2.5 × 10 4 V m 1 ,acting perpendicularly. Calculate the speed of particles perpendicular to electric and ,magnetic field , if their path remains unchanged. The given charge on particles and mass are 3.2 × 10 − 19 C and 12 × 10 − 31 kg , respectively.calculate the radius of circular path traced by the charged particle if the electric field is removed .

Answer

To solve the problem step-by-step, we will first calculate the speed of the charged particles when they enter the region of the electric and magnetic fields, and then we will find the radius of the circular path traced by the charged particle when the electric field is removed.

### Step 1: Understanding the Forces Acting on the Charged Particle

When a charged particle enters a region with both electric and magnetic fields acting perpendicularly, it experiences two forces:
- The electric force (\( F_E \)) due to the electric field, given by:
\[ F_E = qE \]
- The magnetic force (\( F_B \)) due to the magnetic field, given by:
\[ F_B = qvB \]

For the path of the charged particle to remain unchanged, these two forces must be equal in magnitude:
\[ F_E = F_B \]
Thus,
\[ qE = qvB \]

### Step 2: Canceling the Charge and Rearranging the Equation

Since the charge \( q \) appears on both sides of the equation, we can cancel it out (as long as \( q \neq 0 \)):
\[ E = vB \]
Now, we can solve for the speed \( v \):
\[ v = \frac{E}{B} \]

### Step 3: Substituting the Given Values

We have the following values:
- Electric field \( E = 2.5 \times 10^4 \, \text{V/m} \)
- Magnetic field \( B = 5 \times 10^{-3} \, \text{T} \)

Substituting these values into the equation for speed:
\[ v = \frac{2.5 \times 10^4}{5 \times 10^{-3}} \]
Calculating this gives:
\[ v = \frac{2.5 \times 10^4}{5 \times 10^{-3}} = 5.0 \times 10^7 \, \text{m/s} \]

### Step 4: Calculating the Radius of the Circular Path

When the electric field is removed, the charged particle will move in a circular path due to the magnetic force. The centripetal force required for circular motion is provided by the magnetic force:
\[ \frac{mv^2}{r} = qvB \]
Rearranging this to find the radius \( r \):
\[ r = \frac{mv}{qB} \]

### Step 5: Substituting the Known Values

We have:
- Mass \( m = 12 \times 10^{-31} \, \text{kg} \)
- Charge \( q = 3.2 \times 10^{-19} \, \text{C} \)
- Speed \( v = 5.0 \times 10^7 \, \text{m/s} \)
- Magnetic field \( B = 5 \times 10^{-3} \, \text{T} \)

Substituting these values into the radius equation:
\[ r = \frac{(12 \times 10^{-31})(5.0 \times 10^7)}{(3.2 \times 10^{-19})(5 \times 10^{-3})} \]

### Step 6: Performing the Calculation

Calculating the numerator:
\[ 12 \times 10^{-31} \times 5.0 \times 10^7 = 60 \times 10^{-24} = 6.0 \times 10^{-23} \, \text{kg m/s} \]

Calculating the denominator:
\[ 3.2 \times 10^{-19} \times 5 \times 10^{-3} = 16 \times 10^{-22} = 1.6 \times 10^{-21} \, \text{C T} \]

Now substituting back into the radius equation:
\[ r = \frac{6.0 \times 10^{-23}}{1.6 \times 10^{-21}} = 0.0375 \, \text{m} = 3.75 \, \text{cm} \]

### Final Answers
- Speed of the particles: \( v = 5.0 \times 10^7 \, \text{m/s} \)
- Radius of the circular path: \( r = 3.75 \, \text{cm} \)